Please check my work.thank you so much
Quadratic Inequalities are always easiest to solve if you complete the square.
$\displaystyle \displaystyle x^2 - 2x - 3 \leq 0$
$\displaystyle \displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 3 \leq 0$
$\displaystyle \displaystyle (x - 1)^2 - 4 \leq 0$
$\displaystyle \displaystyle (x - 1)^2 \leq 4$
$\displaystyle \displaystyle \sqrt{(x-1)^2} \leq \sqrt{4}$
$\displaystyle \displaystyle |x-1| \leq 2$
$\displaystyle \displaystyle -2 \leq x - 1 \leq 2$
$\displaystyle \displaystyle -1 \leq x \leq 3$.
Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.
For example, to solve $\displaystyle \displaystyle x^2 - 2x - 3 \leq 0$, first solve $\displaystyle \displaystyle x^2 - 2x - 3= 0$, by factoring, $\displaystyle (x- 3)(x+ 1)= 0$ so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it must pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.
To decide which interval is ">0" and which "< 0" you can use either of two methods:
1) Choose a number in each interval and check the sign. x= -2 is less than -1 and $\displaystyle (-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0$ so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and $\displaystyle (0)^2- 2(0)- 3=-3< 0$ so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and $\displaystyle (4)^2- 2(4)- 3= 5> 0$ so the polynomial is greater than 0 for all x greater than 3. Since the inequality say "$\displaystyle \le 0$", the solution set is $\displaystyle -1\le x\le 3$.
2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It.
$\displaystyle x^2-2x-3\ \le\ 0$
$\displaystyle (x-3)(x+1)\le\ 0$
Your Case 1:
If $\displaystyle x-3 \ge\ 0$, then $\displaystyle x+1=(x-3)+4$ must be $\displaystyle \ge\ 0$ since it is 4 greater than (x-3)
No luck there...
Case 2:
If $\displaystyle x-3\ \le\ 0$, then $\displaystyle x\ \le\ 3$ and $\displaystyle x+1\ \ge\ 0$ only for $\displaystyle x\ \ge\ -1$
Hence $\displaystyle -1\ \le\ x\ \le\ 3$
so your answer is correct and the diagrams help illustrate the intersections.
Yet another way to proceed is....
$\displaystyle x^2-2x\ \le\ 3\Rightarrow\ x(x-2)\ \le\ 3$
The factors of 3 that differ by 2 are 1 and 3 or $\displaystyle -1$ and $\displaystyle -3$
$\displaystyle x(x-2)=3$ for $\displaystyle x=3$ and for $\displaystyle x=-1$
If $\displaystyle x<-1$, then $\displaystyle x(x-2)>3$
If $\displaystyle x>3$, then $\displaystyle x(x-2)>3$
The solution is $\displaystyle -1\ \le\ x\ \le\ 3$
$\displaystyle x^2 - 2x - 3 \le 0$
$\displaystyle (x + 1)(x - 3) \le 0$
think about the graph of the quadratic ... a parabola that opens upward with two zeros, $\displaystyle x = -1$ and $\displaystyle x = 3$
the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros.